Find the volume under $z = 3x$ and about he $1$st quadrant area in the $xy$ plane bounded by
$$ x = 0,\quad y = 0,\quad x = 4, \quad x^{2} + y^{2}=25, $$
$$\int_{0}^{4}\int_{0}^{25-x^{2}}3x\,{\rm d}x\,{\rm d}y = 98.$$
Not sure if this is correct. I was confused if this took the volume of the area b/w $$4<x<5 $$ and $$0<y<\sqrt{25-x^{2}}$$ as well, so a second integral would be needed to remove this portion of volume.
I appreciate the help. Thanks
On
I would say, in the second integration substitute $x = 5\sin t, \ dx = 5\cos t\,dt$
$\displaystyle V=3\int_{0}^{4} \int_{0}^{\color{red}{\sqrt{25-x^2}}} x \,dy\, dx=\cdots =3\int_{0}^{4} x\sqrt{25-x^2}\,dx=3\cdot 125\int_{0}^{\arccos \frac{3}{5}}\sin t \cos^2 t \,dt=\\=-125 \cdot [\cos^3 t]_0^{\arccos \frac{3}{5}}=\cdots = 98$
$$Volume= \int_{0}^4 \int_{0}^{\sqrt{25-x^2}} 3x \ dy \ dx $$